Question

Use the definition of taylor series to find the taylor series (centered at

Answer 1

Cos(x) is infinitely differentiable, can be expanded using Taylor's series.
The series about c can be expressed as
$f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(c)}{n!}(x-c)^n$
Substituting cos(x), c= π /4,we have
$cos(x)=\sum_{n=0}^{\infty}\frac{f^{n}(\frac{\pi}{4})}{n!}(x-\frac{\pi}{4})^n$
and since
$f'(\frac{\pi}{4})=-sin(\frac{\pi}{4})=-\sqrt{2}/2$
$f"(\frac{\pi}{4})=-cos(\frac{\pi}{4})=-\sqrt{2}/2$
$f^{(iii)}(\frac{\pi}{4})=sin(\frac{\pi}{4})=\sqrt{2}/2$
$f^{(iv)}(\frac{\pi}{4})=cos(\frac{\pi}{4})=\sqrt{2}/2$

we can simplify the expansion to 
$cos(x)=\frac{\sqrt{2}}{2}(1-(x-\frac{\pi}{4})/1!-(x-\frac{\pi}{4})^2/2!+(x-\frac{\pi}{4})^3/3!+(x-\frac{\pi}{4})^4/4!-(x-\frac{\pi}{4})^5/5!-...)$
Note that the sign pattern is  - - + + - - + + - - ..... following the sign pattern of the derivatives of cos(x).